Multiplicity and set of zeros.

exercise:

Let us assume that the function f has derivatives of all orders.

Suppose that all zeros of $f$ have finite multiplicity. Let $a$ and $b$ be points of $A$, such that $a<b$ and neither point is a zero. Show that $f$ has at most finitely many zeros in $] a, b[$

(We say that a point $c$ is a root of $f(x)=0$ with multiplicity $m$, if $f^{(k)}(c)=0$ for $k=0, \ldots, m-1$ and $f^{(m)}(c) \neq 0 .$ As usual $f^{(0)}$ denotes $f$.)

lemma:

A zero of finite multiplicity is an isolated point of the set of zeros

proof:

Considering Taylor polynomial $E(h)=f(c+h)-\left(f(c)+\frac{1}{1 !} f^{\prime}(c) h+\frac{1}{2 !} f^{\prime \prime}(c) h^{2}+\cdots+\frac{1}{m !} f^{(m)}(c) h^{m}\right)$ and using L’Hopital’s Rule we can show that $\lim\limits_{h \rightarrow 0} \frac{E(h)}{h^{m}}=0$. Because of the multiplicity and for $h \neq 0$ we can write $$ \frac{f(c+h)-f(c)}{h^{m}}=\frac{1}{m !} f^{(m)}(c)+\frac{E(h)}{h^{m}} $$ and from this we can deduce the proof of the lemma.

So if we know if all zeros of $f$ have finite multiplicity then all zeros are isolated. Meanwhile we have bounded interval we can use Bolzano–Weierstrass theorem to show that if f have infinitely many zeros then at least one of the zeros should be limit point which contradicts to be isolated.

But why I do need $f(a)$, $f(b)$ not being zero which was declared in the exercise. I am suspicious about using intermediate value theorem and taylor polynomial with remainder. But I don't know how to do it??

Remark: ( May be it can be some kind of hint which I cannot figure out: next exercise asks ::: In the previous exercise, if f (a) and f (b) have the same sign, show that the number of zeros in ]a, b[, counted by multiplicity, is even. If $f(a)$ and $f(b)$ have opposite signs, show that the number of zeros in ]a, b[, counted by multiplicity, is odd)

Solution 1:

I think it's just to avoid a proliferation of cases in the proof. If you allow $f(a)=0$, for instance, you have to say "$n$ times right-differentiable" every time, instead of just "$n$ times differentiable". A function can have right-derivatives of all orders; for instance,$$f(x)=\begin{cases} e^{-1/x^2}\sin\frac{1}{x}&\text{if }x\ne 0\\ 0&\text{if }x=0 \end{cases}$$

on $[0,1]$. This has an infinite number of zeroes, all of finite multiplicity except $f(0)$, which has infinite multiplicity. But then you have to define what the multiplicity of a zero is at the end-points of your interval. Not difficult, just messy.

Solution 2:

Considering very useful comments and the answer of @TonyK and after long thoughts I have made conclusion:

What we know:

1. All zeros are isolated.

2. Given open bounded interval but the end points is in the domain of the function.

So to find a contradiction we assume that there are infinitely many zeros in $(a,b)$ If that is, we can construct sequence $\left(a_{n}\right)_{n=1}^{\infty}$ where $f(a_{n})=0$. Meanwhile interval is bounded by Bolzano–Weierstrass theorem we have a subsequence $\left(a_{k_{n}}\right)_{n=1}^{\infty}$ which $\lim\limits _{n \rightarrow \infty} a_{k_{n}}=t$, and by continuity we have $f(t)=0$.

If $t=a$ then $f(a)=0$ but it is excluded because of the assumption of the exercise. Similarly $t=b \Rightarrow f(b)=0$ is ruled out.Now $t$ must be interior limit point of the interval $(a,b)$ and again by continuity $f(t)$ should be zero. But as a limit point being zero of the function contradicts with the fact that "all zeros are isolated".

I would be grateful if somebody check my answer.